Mon. Sep 16th, 2024

Dec 9, 2020

Koshe Dekhi 8.5Class 9

Koshe Dekhi 8.5 Class 9

উৎপাদকে বিশ্লেষণ

1. নিচের বহুপদী সংখ্যামালাগুলিকে করি :

(i) $\fn_cm&space;{\color{Blue}&space;\left&space;(&space;a+b&space;\right&space;)^{2}-5a-5b+6}$

সমাধানঃ

$\fn_cm&space;\left&space;(&space;a+b&space;\right&space;)^{2}-5a-5b+6$

$\fn_cm&space;=\left&space;(&space;a+b&space;\right&space;)^{2}-5\left&space;(a+b&space;\right&space;)+6$

$\fn_cm&space;=\left&space;(&space;a+b&space;\right&space;)^{2}-3\left&space;(a+b&space;\right&space;)-2\left&space;(a+b&space;\right&space;)+6$

$=\left&space;(&space;a+b&space;\right&space;)\left&space;[&space;\left&space;(&space;a+b&space;\right&space;)-3&space;\right&space;]-2\left&space;[&space;\left&space;(&space;a+b&space;\right&space;)-3&space;\right&space;]$

$=\left&space;(&space;a+b-3&space;\right&space;)\left&space;(&space;a+b-2&space;\right&space;)$

Koshe dekhi 8.5 class 9

উৎপাদকে বিশ্লেষণ

(ii) $\fn_cm&space;{\color{Blue}&space;\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;x+2&space;\right&space;)\left&space;(&space;3x-1&space;\right&space;)\left&space;(&space;3x-4&space;\right&space;)+12}$

সমাধানঃ

$\fn_cm&space;\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;x+2&space;\right&space;)\left&space;(&space;3x-1&space;\right&space;)\left&space;(&space;3x-4&space;\right&space;)+12$

$=\left&space;[&space;\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;3x-1&space;\right&space;)&space;\right&space;]\left&space;[&space;\left&space;(&space;x+2&space;\right&space;)\left&space;(&space;3x-4&space;\right&space;)&space;\right&space;]+12$

$=\left&space;[&space;3x^{2}-x+3x-1&space;\right&space;]\left&space;[&space;3x^{2}-4x+6x-8&space;\right&space;]+12$

$=\left&space;[&space;3x^{2}+2x-1&space;\right&space;]\left&space;[&space;3x^{2}+2x-8&space;\right&space;]+12$

$=\left&space;(&space;a-1&space;\right&space;)\left&space;(&space;a-8&space;\right&space;)+12$    [ ধরি, ${\color{Blue}&space;3x^{2}+2x=a]}$

$=a^{2}-8a-a+8+12$

$=a^{2}-9a+20$

$=a^{2}-5a-4a+20$

$=a\left&space;(&space;a-5&space;\right&space;)-4\left&space;(&space;a-5&space;\right&space;)$

$=\left&space;(&space;a-5&space;\right&space;)\left&space;(&space;a-4&space;\right&space;)$

$=\left&space;(&space;3x^{2}+2x-5&space;\right&space;)\left&space;(&space;3x^{2}+2x-4&space;\right&space;)$ [ a এর মান বসিয়ে পাই ]

$=\left&space;(&space;3x^{2}+5x-3x-5&space;\right&space;)\left&space;(&space;3x^{2}+2x-4&space;\right&space;)$

$=\left&space;[x\left&space;(&space;3x+5&space;\right&space;)-1\left&space;(&space;3x+5&space;\right&space;)&space;\right&space;]\left&space;(&space;3x^{2}+2x-4&space;\right&space;)$

$=\left&space;(&space;3x+5&space;\right&space;)\left&space;(&space;x-1&space;\right&space;)\left&space;(&space;3x^{2}+2x-4&space;\right&space;)$

Koshe dekhi 8.5 class 9

উৎপাদকে বিশ্লেষণ

(iii) $\fn_cm&space;{\color{Blue}&space;x\left&space;(&space;x^{2}-1&space;\right&space;)\left&space;(&space;x+2&space;\right&space;)-8}$

সমাধানঃ

$\fn_cm&space;x\left&space;(&space;x^{2}-1&space;\right&space;)\left&space;(&space;x+2&space;\right&space;)-8$

$=x\left&space;[\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;x-1&space;\right&space;)&space;\right&space;]\left&space;(&space;x+2&space;\right&space;)-8$

[ সূত্রঃ ${\color{Blue}&space;a^{2}-b^{2}=\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)}$ ]

$=x\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;x-1&space;\right&space;)&space;\left&space;(&space;x+2&space;\right&space;)-8$

$=\left&space;[x\left&space;(&space;x+1&space;\right&space;)&space;\right&space;]\left&space;[\left&space;(&space;x-1&space;\right&space;)&space;\left&space;(&space;x+2&space;\right&space;)&space;\right&space;]-8$

$=\left&space;(&space;x^{2}+x&space;\right&space;)\left&space;(&space;x^{2}+2x-x-2&space;\right&space;)-8$

$=\left&space;(&space;x^{2}+x&space;\right&space;)\left&space;(&space;x^{2}+x-2&space;\right&space;)-8$

$=a\left&space;(&space;a-2&space;\right&space;)-8$ [ ধরি, ${\color{Blue}&space;x^{2}+x=a}$ ]

$=a^{2}-2a-8$

$=a^{2}-4a+2a-8$

$=a\left&space;(&space;a-4&space;\right&space;)+2\left&space;(&space;a-4&space;\right&space;)$

$=\left&space;(&space;a+2&space;\right&space;)\left&space;(&space;a-4&space;\right&space;)$

$=\left&space;(&space;x^{2}+x+2&space;\right&space;)\left&space;(&space;x^{2}+x-4&space;\right&space;)$ [ a এর মান বসিয়ে পাই ]

উৎপাদকে বিশ্লেষণ

(iv) $\fn_cm&space;{\color{Blue}&space;7\left&space;(&space;a^{2}+b^{2}&space;\right&space;)^{2}-15\left&space;(&space;a^{4}-b^{4}&space;\right&space;)+8\left&space;(&space;a^{2}-b^{2}&space;\right&space;)^{2}}$

সমাধানঃ

$\fn_cm&space;7\left&space;(&space;a^{2}+b^{2}&space;\right&space;)^{2}-15\left&space;(&space;a^{4}-b^{4}&space;\right&space;)+8\left&space;(&space;a^{2}-b^{2}&space;\right&space;)^{2}$

$=7\left&space;(&space;a^{2}+b^{2}&space;\right&space;)^{2}-15\left&space;[\left&space;(&space;a^{2}+b^{2}&space;\right&space;)\left&space;(&space;a^{2}-b^{2}&space;\right&space;)&space;\right&space;]+8\left&space;(&space;a^{2}-b^{2}&space;\right&space;)^{2}$

[ সূত্রঃ ${\color{Blue}&space;a^{2}-b^{2}=\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)}$ ]

$=7p^{2}-15pq+8q^{2}$

[ ধরি, ${\color{Blue}&space;\left&space;(&space;a^{2}+b^{2}&space;\right&space;)=p,\left&space;(&space;a^{2}-b^{2}&space;\right&space;)=q}$ ]

$=7p^{2}-7pq-8pq+8q^{2}$

$=7p\left&space;(&space;p-q&space;\right&space;)-8q\left&space;(&space;p-q&space;\right&space;)$

$=\left&space;(&space;p-q&space;\right&space;)\left&space;(&space;7p-8q&space;\right&space;)$

$=\left&space;(&space;a^{2}+b^{2}-a^{2}+b^{2}&space;\right&space;)\left&space;[&space;7\left&space;(&space;a^{2}+b^{2}&space;\right&space;)-8\left&space;(a^{2}-b^{2}&space;\right&space;)&space;\right&space;]$ [ p ও q এর মান বসিয়ে পাই ]

$=2b^{2}\left&space;(&space;7a^{2}+7b^{2}-8a^{2}+8b^{2}&space;\right&space;)$

$=2b^{2}\left&space;(&space;15b^{2}-a^{2}&space;\right&space;)$

উৎপাদকে বিশ্লেষণ

(v) $\fn_cm&space;{\color{Blue}&space;\left&space;(&space;x^{2}-1&space;\right&space;)^{2}+8x\left&space;(&space;x^{2}+1&space;\right&space;)+19x^{2}}$

সমাধানঃ

$\fn_cm&space;\left&space;(&space;x^{2}-1&space;\right&space;)^{2}+8x\left&space;(&space;x^{2}+1&space;\right&space;)+19x^{2}$

$\fn_cm&space;=\left&space;(&space;x^{2}+1&space;\right&space;)^{2}-4x^{2}+8x\left&space;(&space;x^{2}+1&space;\right&space;)+19x^{2}$

${\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a-b&space;\right&space;)^{2}=\left&space;(&space;a+b&space;\right&space;)^{2}-4ab&space;\right&space;]}$

$\fn_cm&space;=\left&space;(&space;x^{2}+1&space;\right&space;)^{2}+8x\left&space;(&space;x^{2}+1&space;\right&space;)+15x^{2}$

$\fn_cm&space;=\left&space;(&space;x^{2}+1&space;\right&space;)^{2}+5x\left&space;(&space;x^{2}+1&space;\right&space;)+3x\left&space;(&space;x^{2}+1&space;\right&space;)+15x^{2}$

$\fn_cm&space;=\left&space;(&space;x^{2}+1&space;\right&space;)\left&space;[\left&space;(&space;x^{2}+1&space;\right&space;)+5x&space;\right&space;]+3x\left&space;[\left&space;(&space;x^{2}+1&space;\right&space;)+5x&space;\right&space;]$

$=\left&space;(&space;x^{2}+1+3x&space;\right&space;)\left&space;(&space;x^{2}+1+5x&space;\right&space;)$

$\fn_cm&space;=\left&space;(&space;x^{2}+3x+1&space;\right&space;)\left&space;(&space;x^{2}+5x+1&space;\right&space;)$

উৎপাদকে বিশ্লেষণ

(vi) $\fn_cm&space;{\color{Blue}&space;\left&space;(&space;a-1&space;\right&space;)x^{2}-x-\left&space;(&space;a-2&space;\right&space;)}$

সমাধানঃ

$\fn_cm&space;\left&space;(&space;a-1&space;\right&space;)x^{2}-x-\left&space;(&space;a-2&space;\right&space;)$

$=ax^{2}-x^{2}-x-a+2$

$=ax^{2}-x^{2}-x-a+1+1$

$=ax^{2}-a-x^{2}+1-x+1$

$=a\left&space;(&space;x^{2}-1&space;\right&space;)-\left&space;(&space;x^{2}-1&space;\right&space;)-1\left&space;(&space;x-1&space;\right&space;)$

$=a\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;x-1&space;\right&space;)-\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;x-1&space;\right&space;)-1\left&space;(&space;x-1&space;\right&space;)$

[ সূত্রঃ ${\color{Blue}&space;a^{2}-b^{2}=\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)}$ ]

$=\left&space;(&space;x-1&space;\right&space;)\left&space;[&space;a\left&space;(&space;x+1&space;\right&space;)-\left&space;(&space;x+1&space;\right&space;)&space;-1\right&space;]$

$=\left&space;(&space;x-1&space;\right&space;)\left&space;(&space;ax+a-x-1-1&space;\right&space;)$

$=\left&space;(&space;x-1&space;\right&space;)\left&space;(&space;ax+a-x-2&space;\right&space;)$

উৎপাদকে বিশ্লেষণ

(vii) $\fn_cm&space;{\color{Blue}&space;\left&space;(&space;a-1&space;\right&space;)x^{2}+a^{2}xy+\left&space;(&space;a+1&space;\right&space;)y^{2}}$

সমাধানঃ

$\left&space;(&space;a-1&space;\right&space;)x^{2}+a^{2}xy+\left&space;(&space;a+1&space;\right&space;)y^{2}$

$=\left&space;(&space;a-1&space;\right&space;)x^{2}+\left&space;[&space;\left&space;(&space;a^{2}-1&space;\right&space;)+1&space;\right&space;]xy+\left&space;(&space;a+1&space;\right&space;)y^{2}$

$=\left&space;(&space;a-1&space;\right&space;)x^{2}+&space;\left&space;(&space;a^{2}-1&space;\right&space;)xy+xy+\left&space;(&space;a+1&space;\right&space;)y^{2}$

$=\left&space;(&space;a-1&space;\right&space;)x\left&space;[&space;x+\left&space;(&space;a+1&space;\right&space;)y&space;\right&space;]+y\left&space;[&space;x+\left&space;(&space;a+1&space;\right&space;)y&space;\right&space;]$

$=\left&space;(&space;ax-x&space;\right&space;)\left&space;[&space;x+ay+y&space;\right&space;]+y\left&space;[&space;x+ay+y&space;\right&space;]$

$=\left&space;(&space;x+ay+y&space;\right&space;)\left&space;(&space;ax-x+y&space;\right&space;)$

Koshe dekhi 8.5 class 9

উৎপাদকে বিশ্লেষণ

(viii) $\fn_cm&space;{\color{Blue}&space;x^{2}-qx-p^{2}+5pq-6q^{2}}$

সমাধানঃ

$\fn_cm&space;x^{2}-qx-p^{2}+5pq-6q^{2}$

$\fn_cm&space;=x^{2}-qx-\left&space;(p^{2}-5pq+6q^{2}&space;\right&space;)$

$\fn_cm&space;=x^{2}-qx-\left&space;(p^{2}-3pq-2pq+6q^{2}&space;\right&space;)$

$\fn_cm&space;=x^{2}-qx-\left&space;[p\left&space;(p-3q&space;\right&space;)-2q\left&space;(&space;p-3q&space;\right&space;)&space;\right&space;]$

$\fn_cm&space;=x^{2}-qx-\left&space;(p-3q&space;\right&space;)\left&space;(&space;p-2q&space;\right&space;)$

$\fn_cm&space;=x^{2}-\left&space;[&space;\left&space;(&space;p-2q&space;\right&space;)-\left&space;(&space;p-3q&space;\right&space;)&space;\right&space;]x-\left&space;(p-3q&space;\right&space;)\left&space;(&space;p-2q&space;\right&space;)$

$\fn_cm&space;=x^{2}-\left&space;(&space;p-2q&space;\right&space;)x+\left&space;(&space;p-3q&space;\right&space;)x-\left&space;(p-3q&space;\right&space;)\left&space;(&space;p-2q&space;\right&space;)$

$=x\left&space;(&space;x-p+2q&space;\right&space;)+\left&space;(&space;p-3q&space;\right&space;)\left&space;(&space;x-p+2q&space;\right&space;)$

$=\left&space;(&space;x+p-3q&space;\right&space;)\left&space;(&space;x-p+2q&space;\right&space;)$

উৎপাদকে বিশ্লেষণ

(ix) $\fn_cm&space;{\color{Blue}&space;2\left&space;(&space;a^{2}+\frac{1}{a^{2}}&space;\right&space;)-\left&space;(&space;a-\frac{1}{a}&space;\right&space;)-7}$

সমাধানঃ

$\fn_cm&space;{\color{Blue}&space;2\left&space;(&space;a^{2}+\frac{1}{a^{2}}&space;\right&space;)-\left&space;(&space;a-\frac{1}{a}&space;\right&space;)-7}$

$\fn_cm&space;=2\left&space;[\left&space;(&space;a-\frac{1}{a}&space;\right&space;)^{2}+2&space;\right&space;]-\left&space;(&space;a-\frac{1}{a}&space;\right&space;)-7$

ধরি, ${\color{Blue}&space;\left&space;(&space;a-\frac{1}{a}&space;\right&space;)=x}$

$\fn_cm&space;=2\left&space;(&space;x^{2}+2&space;\right&space;)&space;-x-7$

$\fn_cm&space;=2x^{2}+4-x-7$

$\fn_cm&space;=2x^{2}-x-3$

$\fn_cm&space;=2x^{2}-3x+2x-3$

$=x\left&space;(&space;2x-3&space;\right&space;)+1\left&space;(&space;2x-3&space;\right&space;)$

$=\left&space;(&space;2x-3&space;\right&space;)\left&space;(&space;x+1&space;\right&space;)$

x এর মান বসিয়ে পাই,

$=\left&space;[&space;2\left&space;(&space;a-\frac{1}{a}-3&space;\right&space;)&space;\right&space;]\left&space;(&space;a-\frac{1}{a}+1&space;\right&space;)$

$=\left&space;(&space;2a-\frac{2}{a}-3&space;\right&space;)\left&space;(&space;a-\frac{1}{a}+1&space;\right&space;)$

উৎপাদকে বিশ্লেষণ

(x) $\fn_cm&space;{\color{Blue}&space;\left&space;(&space;x^{2}-x&space;\right&space;)y^{2}+y-\left&space;(&space;x^{2}+x&space;\right&space;)}$

সমাধানঃ

$\fn_cm&space;\left&space;(&space;x^{2}-x&space;\right&space;)y^{2}+y-\left&space;(&space;x^{2}+x&space;\right&space;)$

$\fn_cm&space;=x\left&space;(&space;x-1&space;\right&space;)y^{2}+y-x\left&space;(&space;x+1&space;\right&space;)$

$=x\left&space;(&space;x-1&space;\right&space;)y^{2}+\left&space;[&space;x^{2}-\left&space;\{&space;\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;x-1&space;\right&space;)&space;\right&space;\}&space;\right&space;]y-x\left&space;(&space;x+1&space;\right&space;)$

$=x\left&space;(&space;x-1&space;\right&space;)y^{2}+x^{2}y-\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;x-1&space;\right&space;)&space;y-x\left&space;(&space;x+1&space;\right&space;)$

$=xy\left&space;[&space;\left&space;(&space;x-1&space;\right&space;)y+x&space;\right&space;]-\left&space;(&space;x+1&space;\right&space;)\left&space;[&space;\left&space;(&space;x-1&space;\right&space;)y+x&space;\right&space;]$

$=\left&space;[&space;\left&space;(&space;x-1&space;\right&space;)y+x&space;\right&space;]\left&space;[xy-\left&space;(&space;x+1&space;\right&space;)&space;\right&space;]$

$=\left&space;(&space;xy-y+x&space;\right&space;)\left&space;(&space;xy-x-1&space;\right&space;)$

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(i) $\fn_cm&space;{\color{Blue}&space;a^{2}-b^{2}=11\times&space;9}$ এবং a ও b ধনাত্মক পূর্ণসংখ্যা (a > b) হলে

(a) a = 11, b = 9

(b) a = 33, b = 3

(c) a = 10, b = 1

(d) a = 100, b = 1

সমাধানঃ

$\fn_cm&space;a^{2}-b^{2}=11\times&space;9$

বা, $\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=\left&space;(&space;10+1&space;\right&space;)\left&space;(&space;10-1&space;\right&space;)$

উভয়পক্ষ তুলনা করে পাই,

$a=10,b=1$

উত্তরঃ (c) a = 10, b = 1

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(ii) যদি  $\fn_cm&space;{\color{Blue}&space;\frac{a}{b}+&space;\frac{b}{a}=1}$ হয়, তাহলে $\fn_cm&space;{\color{Blue}&space;a^{3}+b^{3}}$ -এর মান

(a) 1

(b) a

(c) b

(d) 0

সমাধানঃ

$\fn_cm&space;\frac{a}{b}+&space;\frac{b}{a}=1$

বা, $\frac{a^{2}+b^{2}}{ab}=1$

বা, $a^{2}+b^{2}=ab$

বা, $a^{2}+b^{2}-ab=0$

বা, $\left&space;(&space;a+b&space;\right&space;)\left&space;(a^{2}+b^{2}-ab&space;\right&space;)=0\times&space;\left&space;(&space;a+b&space;\right&space;)$

বা, $a^{3}+b^{3}=0$

উত্তরঃ (d) 0

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(iii) $\fn_cm&space;{\color{Blue}&space;25^{3}-75^{3}+50^{3}+3\times&space;25\times&space;75\times&space;50}$ -এর মান

(a) 150

(b) 0

(c) 25

(d) 50

সমাধানঃ

$\fn_cm&space;25^{3}-75^{3}+50^{3}+3\times&space;25\times&space;75\times&space;50$

$=\left&space;(&space;25-75+50&space;\right&space;)\left&space;[&space;\left&space;(&space;25&space;\right&space;)^{2}+\left&space;(&space;-75&space;\right&space;)^{2}+\left&space;(&space;50&space;\right&space;)^{2}-25\times&space;\left&space;(&space;-75&space;\right&space;)-\left&space;(&space;-75&space;\right&space;)\times&space;50-50\times&space;25&space;\right&space;]$

${\color{Blue}&space;\left&space;[&space;\because&space;a^{3}+b^{3}+c^{3}+3abc=\left&space;(&space;a+b+c&space;\right&space;)\left&space;(&space;a^{2}+b^{2}+c^{2}-ab-bc-ca&space;\right&space;)\right&space;]}$

= 0

উত্তরঃ (b) 0

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(iv) $\fn_cm&space;{\color{Blue}&space;a+b+c=0}$ হলে $\fn_cm&space;{\color{Blue}&space;\frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}}$ -এর মান

(a) 0

(b) 1

(c) −1

(d) 3

সমাধানঃ

$\fn_cm&space;\frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}$

$=\frac{a^{3}+b^{3}+c^{3}}{abc}$

$=\frac{\left&space;(&space;a+b+c&space;\right&space;)\left&space;(&space;a^{2}+b^{2}+c^{2}-ab-bc-ca&space;\right&space;)+3abc}{abc}$

$=\frac{0\times&space;\left&space;(&space;a^{2}+b^{2}+c^{2}-ab-bc-ca&space;\right&space;)}{abc}+\frac{3abc}{abc}$

= 0 + 3

= 3

উত্তরঃ (d) 3

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(v) $\fn_cm&space;{\color{Blue}&space;x^{2}-px+12=\left&space;(&space;x-3&space;\right&space;)\left&space;(&space;x-a&space;\right&space;)}$ একটি অভেদ হলে, a ও p -এর মান যথাক্রমে

(a) a = 4, p = 7

(b) a = 7, p = 4

(c) a = 4, p = −7

(d) a = −4, p = 7

সমাধানঃ

$\fn_cm&space;x^{2}-px+12=\left&space;(&space;x-3&space;\right&space;)\left&space;(&space;x-a&space;\right&space;)$

বা, $\fn_cm&space;x^{2}-px+12=x^{2}-3x-ax+3a$

বা, $\fn_cm&space;x^{2}-px+12=x^{2}-x\left&space;(3+a&space;\right&space;)+3a$

উভয়পক্ষ তুলনা করে পাই,

$p=3+a,3a=12$

$\therefore&space;a=\frac{12}{3}=4$

$p=3+4=7$

উত্তরঃ (a) a = 4, p = 7

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

(i) $\fn_cm&space;{\color{Blue}&space;\frac{\left&space;(&space;b^{2}-c^{2}&space;\right&space;)^{3}+\left&space;(&space;c^{2}-a^{2}&space;\right&space;)^{3}+\left&space;(&space;a^{2}-b^{2}&space;\right&space;)^{3}}{\left&space;(&space;b-c&space;\right&space;)^{3}+\left&space;(&space;c-a&space;\right&space;)^{3}+\left&space;(&space;a-b&space;\right&space;)^{3}}}$ -এর সরলতম মান লিখি।

সমাধানঃ

$\fn_cm&space;\frac{\left&space;(&space;b^{2}-c^{2}&space;\right&space;)^{3}+\left&space;(&space;c^{2}-a^{2}&space;\right&space;)^{3}+\left&space;(&space;a^{2}-b^{2}&space;\right&space;)^{3}}{\left&space;(&space;b-c&space;\right&space;)^{3}+\left&space;(&space;c-a&space;\right&space;)^{3}+\left&space;(&space;a-b&space;\right&space;)^{3}}$

$=\frac{\left&space;(&space;b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2}&space;\right&space;)\left&space;[\left&space;(b^{2}-c^{2}&space;\right&space;)^{2}+\left&space;(c^{2}-a^{2}&space;\right&space;)^{2}+\left&space;(a^{2}-b^{2}&space;\right&space;)^{2}-\left&space;(b^{2}-c^{2}&space;\right&space;)^{2}\left&space;(c^{2}-a^{2}&space;\right&space;)-\left&space;(c^{2}-a^{2}&space;\right&space;)\left&space;(a^{2}-b^{2}&space;\right&space;)-\left&space;(a^{2}-b^{2}&space;\right&space;)\left&space;(b^{2}-c^{2}&space;\right&space;)&space;\right&space;]+3\left&space;(&space;b^{2}-c^{2}&space;\right&space;)\left&space;(c^{2}-a^{2}&space;\right&space;)\left&space;(a^{2}-b^{2}&space;\right&space;)}{\left&space;(&space;b-c+c-a+a-b&space;\right&space;)\left&space;[&space;\left&space;(&space;b-c&space;\right&space;)^{2}+\left&space;(&space;c-a&space;\right&space;)^{2}+\left&space;(&space;a-b&space;\right&space;)^{2}-\left&space;(&space;b-c&space;\right&space;)\left&space;(&space;c-a&space;\right&space;)-\left&space;(&space;c-a&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)-\left&space;(&space;a-b&space;\right&space;)&space;\left&space;(&space;b-c&space;\right&space;)\right&space;]+3\left&space;(&space;b-c&space;\right&space;)\left&space;(&space;c-a&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)}$

${\color{Blue}&space;\left&space;[&space;\because&space;x^{3}+y^{3}+z^{3}=\left&space;(&space;x+y+z&space;\right&space;)\left&space;(&space;x^{2}+y^{2}+z^{2}-xy-yz-zx&space;\right&space;)+3xyz&space;\right&space;]}$

$=\frac&space;{3\left&space;(&space;b^{2}-c^{2}&space;\right&space;)\left&space;(c^{2}-a^{2}&space;\right&space;)\left&space;(a^{2}-b^{2}&space;\right&space;)}{3\left&space;(&space;b-c&space;\right&space;)\left&space;(&space;c-a&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)}$

$=\left&space;(&space;b+c&space;\right&space;)\left&space;(&space;c+a&space;\right&space;)\left&space;(&space;a+b&space;\right&space;)$

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

(ii) $\fn_cm&space;{\color{Blue}&space;a^{3}+b^{3}+c^{3}-3abc=0}$ এবং $\fn_cm&space;{\color{Blue}&space;a+b+c\neq&space;0}$ হলে a, b ও c -এর মধ্যে সম্পর্ক লিখি।

সমাধানঃ

$\fn_cm&space;a^{3}+b^{3}+c^{3}-3abc=0$

বা, $\left&space;(&space;a+b+c&space;\right&space;)\left&space;(&space;a^{2}+b^{2}+c^{2}-ab-bc-ca&space;\right&space;)=0$

$\fn_cm&space;\because&space;a+b+c\neq&space;0$

$\therefore&space;\left&space;(&space;a^{2}+b^{2}+c^{2}-ab-bc-ca&space;\right&space;)=0$

বা, $2\left&space;(&space;a^{2}+b^{2}+c^{2}-ab-bc-ca&space;\right&space;)=0\times&space;2$

বা, $\left&space;(&space;2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca&space;\right&space;)=0$

বা, $a^{2}-2ab+b^{2}+b^{2}-2bc+c^{2}+c^{2}-2ca+a^{2}=0$

বা, $\left&space;(&space;a-b&space;\right&space;)^{2}+\left&space;(&space;b-c&space;\right&space;)^{2}+\left&space;(&space;c-a&space;\right&space;)^{2}=0$

আমরা জানি, দুই বা ততোধিক রাশির বর্গের সমষ্টি শূন্য হলে তারা প্রত্যেকে পৃথক পৃথক ভাবে শূন্য হয়।

$\therefore&space;\left&space;(&space;a-b&space;\right&space;)^{2}=0$

বা, $a-b=0$

বা, $a=b$

আবার,

$\left&space;(&space;b-c&space;\right&space;)^{2}=0$

বা, $b-c=0$

বা, $b=c$

আবারও

$\left&space;(&space;c-a&space;\right&space;)^{2}=0$

বা, $c-a=0$

বা, $c=a$

${\color{DarkGreen}&space;\therefore&space;a=b=c}$ ইহায় a, b ও c -এর মধ্যে সম্পর্ক।

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

(iii) $\fn_cm&space;{\color{Blue}&space;a^{2}-b^{2}=224}$ এবং a ও b ঋণাত্মক পূর্ণসংখ্যা হলে (a < b) a ও b -এর মান লিখি।

সমাধানঃ

$\fn_cm&space;a^{2}-b^{2}=224$

বা, $\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=\left&space;(&space;-14&space;\right&space;)\times&space;\left&space;(&space;-16&space;\right&space;)$

বা, $\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=\left&space;(&space;-15+1&space;\right&space;)\times&space;\left&space;(&space;-15-1&space;\right&space;)$

উভয়পক্ষ তুলনা করে পাই,

$a=-15,b=-1$

উত্তরঃ নির্ণেয় a ও b এর মান যথাক্রমে -15 ও -1

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

(iv) $\fn_cm&space;{\color{Blue}&space;3x=a+b+c}$  হলে, $\fn_cm&space;{\color{Blue}&space;\left&space;(&space;x-a&space;\right&space;)^{3}+\left&space;(&space;x-b&space;\right&space;)^{3}+\left&space;(&space;x-c&space;\right&space;)^{3}-3\left&space;(&space;x-a&space;\right&space;)\left&space;(&space;x-b&space;\right&space;)\left&space;(&space;x-c&space;\right&space;)}$ -এর মান কত লিখি।

সমাধানঃ

$\fn_cm&space;\left&space;(&space;x-a&space;\right&space;)^{3}+\left&space;(&space;x-b&space;\right&space;)^{3}+\left&space;(&space;x-c&space;\right&space;)^{3}-3\left&space;(&space;x-a&space;\right&space;)\left&space;(&space;x-b&space;\right&space;)\left&space;(&space;x-c&space;\right&space;)$

ধরি,

x – a = p , x – b = q ও x – c = r

এখন,

p + q + r = x – a + x – b + x – c

বা, p + q + r = 3x – a – b – c

বা, p + q + r = 3x – 3x  [ যেহেতু, $\fn_cm&space;{\color{Blue}&space;3x=a+b+c}$ ]

∴ p + q + r = 0

এখন,

$\fn_cm&space;\left&space;(&space;x-a&space;\right&space;)^{3}+\left&space;(&space;x-b&space;\right&space;)^{3}+\left&space;(&space;x-c&space;\right&space;)^{3}-3\left&space;(&space;x-a&space;\right&space;)\left&space;(&space;x-b&space;\right&space;)\left&space;(&space;x-c&space;\right&space;)$

$=p^{3}+q^{3}+r^{3}-3pqr$

$=\left&space;(&space;p+q+r&space;\right&space;)\left&space;(&space;p^{2}+q^{2}+r^{2}-pq-qr-rp&space;\right&space;)+3pqr-3pqr$

$=0\times&space;\left&space;(&space;p^{2}+q^{2}+r^{2}-pq-qr-rp&space;\right&space;)$

= 0

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

(v) $\fn_cm&space;{\color{Blue}&space;2x^{2}+px+6=\left&space;(&space;2x-a&space;\right&space;)\left&space;(&space;x-2&space;\right&space;)}$ একটি অভেদ হলে a ও p -এর মান কত লিখি।

সমাধানঃ

$\fn_cm&space;2x^{2}+px+6=\left&space;(&space;2x-a&space;\right&space;)\left&space;(&space;x-2&space;\right&space;)$

বা, $\fn_cm&space;2x^{2}+px+6=2x^{2}-4x-ax+2a$

বা, $\fn_cm&space;2x^{2}+px+6=2x^{2}-\left&space;(4+a&space;\right&space;)x+2a$

উভয়পক্ষ তুলনা করে পাই,

$p=-\left&space;(&space;4+a&space;\right&space;)$ এবং $6=2a$

$\therefore&space;a=\frac{6}{2}=3$

এবং

$p=-\left&space;(&space;4+a&space;\right&space;)$

বা, $p=-\left&space;(&space;4+3&space;\right&space;)$ [ a এর মান বসিয়ে পাই ]

$\therefore&space;p=-7$

উত্তরঃ নির্ণেয় a ও p এর মান যথাক্রমে 3 ও -7

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CIT-002 Solved Assignment 2024-25

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